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As the title of the question susggests, I would like to show that

The "trivial bound is that"

\begin{align*} \lim_{Q\to\infty}\lim_{N\to\infty}\frac{1}{N\pi(Q)}\sum_{n<N}\left|\sum_{\substack{p<Q\\p|n}}p-\pi(Q)\right|&\leq \lim_{Q\to\infty}\lim_{N\to\infty}\frac{1}{N\pi(Q)}\sum_{n<N}\left(\sum_{\substack{p<Q\\p|n}}p+\pi(Q)\right)\\ &=2 \end{align*}

where the equality is obtained by noting that $\frac{1}{p}$ numbers are multiplies of $p$, so the expected value of $\sum_{\substack{p<q \\ p|n}}p$ is exactly $\sum_{p<Q}1=\pi(Q)$. Thus, we are looking only for a "$o(\cdot)$" improvement. The first thought of mine would be to note that since $\sum_{\substack{p<q \\p|n}}p$ is an additive function and so by the Turan-Kubilius inequality

$$\sum_{n<N}\left|\sum_{\substack{p<Q\\p|n}}p-\pi(Q)\right|^2\leq 4N\sum_{p<Q}p$$

The issue is, this inequality is worse than the trivial one since apply Cauchy-Shwartz we get that this inequality yields

\begin{align*} \frac{1}{N\pi(Q)}\sum_{n<N}\left|\sum_{\substack{p<Q\\p|n}}p-\pi(Q)\right|&\leq \frac{1}{\pi(Q)}\sqrt{\frac{1}{N}\sum_{n<N}\left|\sum_{\substack{p<Q\\p|n}}p-\pi(Q)\right|^2}\\ &\leq \frac{2}{\pi(Q)}\sqrt{\sum_{p<Q}p}\\ \end{align*}

where by the PNT this last term is on the order of $\sqrt{\frac{Q^2}{\log(Q)}}=\frac{Q}{\sqrt{\log(Q)}}$. This inequality would indicate to us that the sum would $\mathit{diverge}$, namely

$$\lim_{N\to\infty}\frac{1}{N\pi(Q)}\sum_{n<N}\left|\sum_{\substack{p<Q\\p|n}}p-\pi(Q)\right|=\Omega(\sqrt{\log(Q)})$$

I have proved that the sum can't go to zero too fast, and specifically that

$$\lim_{N\to\infty}\frac{1}{N\pi(Q)}\sum_{n<N}\left|\sum_{\substack{p<Q\\p|n}}p-\pi(Q)\right|=\Omega(\log(Q)^{-\epsilon})$$

for any $\epsilon>0$. It does, however, feel natural that this sum should be at least $o(1)$.

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  • $\begingroup$ I think the third equation should read: $\cdots \leq \frac{1}{\sqrt{N}\pi(Q)} \sqrt{\frac{1}{N}\cdots}$ which implies that the limit as $N$ tends to infinity is $0$. $\endgroup$ – David Tweedle Oct 28 '20 at 2:21
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    $\begingroup$ @DavidTweedle Are you sure that you accounted for the extra factor of $N$ that appears in the square root when you apply Cauchy-Shwartz? $\endgroup$ – Milo Moses Oct 28 '20 at 2:28
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    $\begingroup$ The inner limit, $\lim_{N\to\infty}\frac 1{N\pi(Q)}\sum_{n<N}\left|\sum_{p<Q,\,p|n} p-\pi(Q)\right|$ is exactly $\frac1{M\pi(Q)}\sum_{n<M}\left|\sum_{p<Q,\,p|n} p-\pi(Q)\right|$, where $M$ is the product of all primes less than $Q$ since the summand in absolute values is $M$-periodic in $n$. $\endgroup$ – Anthony Quas Oct 28 '20 at 7:37
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    $\begingroup$ If $n$ is uniformly distributed on the set $\{1,\ldots,M\}$, then the events $p|n$ are mutually independent for all $p<Q$. This means you can rewrite the inner limit as $\frac 1{\pi(Q)}\mathbb E\left|\sum_{p<Q} X_p-\pi(Q)\right|$, where the $X_p$'s are independent random variables taking the value $p$ with probability $\frac 1p$ and 0 otherwise. $\endgroup$ – Anthony Quas Oct 28 '20 at 7:49
  • $\begingroup$ @MiloMoses Yes, I made a mistake. Sorry about that! $\endgroup$ – David Tweedle Nov 10 '20 at 13:19
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It seems that the conjecture is false, if I did not miss some asymptotc issue The essence of what follows is that I show that the interior limit exceeds any function of $Q$ tending to $0$.

For any $t\geq 1$, denote by $p_t(Q)$ the density of those $n$ divisible by at least one $p$ with $t\pi(Q)<p<Q$. We have \begin{align*} p_t(Q)&=1-\prod_{t\pi(Q)<p<Q}\left(1-\frac1p\right) \sim 1-\exp\left(\sum_{t\pi(Q)<p<Q}\frac1p\right) \sim 1-\frac{\log Q}{\log (t\pi(Q))}\\ &\sim 1-\frac{\log Q}{\log Q+\log t-\log\log Q} =\frac{\log\log Q-\log t}{\log Q+\log t-\log\log Q}\\ &\sim \frac{\log\log Q-\log t}{\log Q}, \end{align*} where the estimates work in the regime $t\leq o(Q/\pi(Q))=o(\log Q)$. (More precisely, in any such regime this equivalence is uniform over $t\leq o(\log Q)$, as $q\to\infty$.)

Assume now that your conjecture is true, i.e., that $$ f(Q):=\lim_{N\to\infty}\frac{1}{N\pi(Q)}\sum_{n<N}\left|\sum_{\substack{p<Q\\p|n}}p-\pi(Q)\right| \to 0, \qquad Q\to\infty. $$ Take $N$ to be a multiple of $Q!$. Notice that at least $p_t(Q)N$ of the summands are larger than $(t-1)\pi(Q)$. Put $q=\log Q$. Summing up over $t=2,3,\dots,qf(Q)=o(q)$, we obtain \begin{align*} \frac1{N\pi(Q)}\sum_{n<N}\left|\sum_{\substack{p<Q\\p|n}}p-\pi(Q)\right| &\buildrel(*)\over\ge \sum_{t=2}^{qf(Q)} p_t(Q) \sim \sum_{t=2}^{qf(Q)} \frac{\log q-\log t}{q} =\frac1q\log\frac{q^{qf(Q)}}{(qf(Q))!}\\ &\sim \frac1q\log\left(\frac{qe}{qf(Q)}\right)^{qf(Q)} \sim f(Q)(-\log f(Q)). \end{align*} Since $f(Q)=o(1)$, the limit of the above expression as $N\to\infty$ cannot equal $f(Q)$.

Remark. The inequality $(*)$ holds, because if a number $n$ is accounted for $x$ times in the right sum, then its large prime divisor $p$ is larger than $(x+1)\pi(Q)$, so that $$ \left|\sum_{\substack{p<Q\\p|n}}p-\pi(Q)\right|>x\pi(Q). $$

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  • $\begingroup$ Are you sure that there are no issues with counting numbers multiple times in the sum $\sum_{t=2}^{qf(Q)}p_t(Q)$? If it is the density of primes where at least on prime divides in that in those intervals, numbers which appear in the tightest interval will also appear in every other term. $\endgroup$ – Milo Moses Oct 28 '20 at 15:50
  • $\begingroup$ It seems that I am; notice that there is no $\sim$ sign there. I've added a clarification in the Remark at the end. If you are unsure about some other steps --- feel free to ask, this is really sketchy; sorry for that. $\endgroup$ – Ilya Bogdanov Oct 28 '20 at 16:24
  • $\begingroup$ Could you expand the last step in the first line of math where you claim that $1-\frac{\log(Q)}{\log(t\pi(Q))}$ is asymptotic to a much simpler expression (and at a uniform rate) $\endgroup$ – Milo Moses Oct 28 '20 at 16:37
  • $\begingroup$ I've inserted a line into the equation. Is it really the place you got troubls, or perhaps you meany a different one? $\endgroup$ – Ilya Bogdanov Oct 29 '20 at 6:03
  • $\begingroup$ Though it pains me to say it, I can't find any more questionable steps in your proof. Thank you very much for your help. $\endgroup$ – Milo Moses Oct 29 '20 at 15:13

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